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Alexxx [7]
2 years ago
14

Digital learning can help students who enjoy

Computers and Technology
1 answer:
mr Goodwill [35]2 years ago
6 0

Answer:

flexibility and independence

Explanation:

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Answer:

you should talk to them when u think its the right time and you think u can trust them

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3 years ago
Develop a C program that calculates the final score and the average score for a student from his/her (1)class participation, (2)
Ghella [55]

Answer:

#include <iomanip>

#include<iostream>

using namespace std;

int main(){

char name[100];

float classp, test, assgn, exam, prctscore,ave;

cout<<"Student Name: ";

cin.getline(name,100);

cout<<"Class Participation: "; cin>>classp;

while(classp <0 || classp > 100){  cout<<"Class Participation: "; cin>>classp; }

cout<<"Test: "; cin>>test;

while(test <0 || test > 100){  cout<<"Test: "; cin>>test; }

cout<<"Assignment: "; cin>>assgn;

while(assgn <0 || assgn > 100){  cout<<"Assignment: "; cin>>assgn; }

cout<<"Examination: "; cin>>exam;

while(exam <0 || exam > 100){  cout<<"Examination: "; cin>>exam; }

cout<<"Practice Score: "; cin>>prctscore;

while(prctscore <0 || prctscore > 100){  cout<<"Practice Score: "; cin>>prctscore; }

ave = (int)(classp + test + assgn + exam + prctscore)/5;

cout <<setprecision(1)<<fixed<<"The average score is "<<ave;  

return 0;}

Explanation:

The required parameters such as cin, cout, etc. implies that the program is to be written in C++ (not C).

So, I answered the program using C++.

Line by line explanation is as follows;

This declares name as character of maximum size of 100 characters

char name[100];

This declares the grading items as float

float classp, test, assgn, exam, prctscore,ave;

This prompts the user for student name

cout<<"Student Name: ";

This gets the student name using getline

cin.getline(name,100);

This prompts the user for class participation. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

<em> cout<<"Class Participation: "; cin>>classp; </em>

<em> while(classp <0 || classp > 100){  cout<<"Class Participation: "; cin>>classp; } </em>

This prompts the user for test. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

<em> cout<<"Test: "; cin>>test; </em>

<em> while(test <0 || test > 100){  cout<<"Test: "; cin>>test; } </em>

This prompts the user for assignment. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

<em> cout<<"Assignment: "; cin>>assgn; </em>

<em> while(assgn <0 || assgn > 100){  cout<<"Assignment: "; cin>>assgn; } </em>

This prompts the user for examination. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

<em> cout<<"Examination: "; cin>>exam; </em>

<em> while(exam <0 || exam > 100){  cout<<"Examination: "; cin>>exam; } </em>

This prompts the user for practice score. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

<em> cout<<"Practice Score: "; cin>>prctscore; </em>

<em> while(prctscore <0 || prctscore > 100){  cout<<"Practice Score: "; cin>>prctscore; } </em>

This calculates the average of the grading items

ave = (int)(classp + test + assgn + exam + prctscore)/5;

This prints the calculated average

cout <<setprecision(1)<<fixed<<"The average score is "<<ave;  

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kykrilka [37]

Answer:

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1. Attributes

2. Datatypes

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To minimize this problem  

1. We switch low pass L-Network to high pass L-network

2. We switch high pass L-Network to low pass L-network

3. We use the impedence matching transformer

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